The edges of a parallelopiped are of unit length and are parallel to non-coplanar unit vectors a, b,

Q: Why can't we just calculate the base area and multiply by the height? It seems like a standard geometry problem.

That approach works perfectly if you have coordinates, but here we are in the abstract space of vectors. We don't know the height! The Gramian determinant is the bridge that lets us bypass the need for explicit coordinates. It is the most elegant way to handle volume when only the 'relationships' (dot products) between edges are known.

Q: Is the Gramian determinant really necessary? It feels like a shortcut.

It is not just a shortcut; it is a fundamental property of vector spaces! In JEE Advanced, you will often encounter problems where coordinates are hidden. Mastering the Gramian is your superpower. It transforms a difficult geometric visualization problem into a straightforward algebraic one.

Q: Wait, why is the determinant of the dot products equal to the volume squared?

Think of the scalar triple product $[\hat{a} \hat{b} \hat{c}]$ as the signed volume. The Gramian matrix $G$ is defined as $G_{ij} = \vec{v}_i \cdot \vec{v}_j$. It is a beautiful theorem in linear algebra that $\det(G) = |[\hat{a} \hat{b} \hat{c}]|^2$. It essentially measures how 'independent' or 'orthogonal' the vectors are. If they were coplanar, the determinant would be zero!

Question

The edges of a parallelopiped are of unit length and are parallel to non-coplanar unit vectors $\hat{a}, \hat{b}, \hat{c}$ such that $\hat{a} \cdot \hat{b} = \hat{b} \cdot \hat{c} = \hat{c} \cdot \hat{a} = \frac{1}{2}$. Then, the volume of the parallelopiped is

Options:

(A) $\frac{1}{\sqrt{2}}$

(B) $\frac{1}{2\sqrt{2}}$

(C) $\frac{\sqrt{3}}{2}$

(D) $\frac{1}{\sqrt{3}}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, my dear student. Today, we are not just solving a problem; we are peeling back the layers of 3D geometry. Imagine you are standing in a vast, empty space with three unit vectors, $\hat{a}, \hat{b},$ and $\hat{c}$.

These vectors are the foundational edges of a parallelepiped, but they are skewed rather than sitting at right angles. We are given the dot products: $\hat{a} \cdot \hat{b} = \hat{b} \cdot \hat{c} = \hat{c} \cdot \hat{a} = \frac{1}{2}$.

Since these are unit vectors, we know that $\hat{a} \cdot \hat{b} = |\hat{a}||\hat{b}| \cos 	heta = 1 \cdot 1 \cdot \cos 	heta = \frac{1}{2}$. This immediately tells us that the angle between any two of these vectors is $60^\circ$. This is a highly symmetric, beautiful structure.

### The Power of the Gramian Determinant

In the JEE Advanced arena, you will often face problems where the 'obvious' path—finding coordinates—is blocked. We need a tool that works with the relationships between vectors rather than their positions. Enter the **Gramian Determinant**.

The volume $V$ of a parallelepiped defined by vectors $\vec{u}, \vec{v}, \vec{w}$ is given by the absolute value of the scalar triple product: $V = |[\vec{u} \vec{v} \vec{w}]|$. The Gramian identity states that the square of this scalar triple product is equal to the determinant of the matrix of all possible dot products:

$$
V^2 = \begin{vmatrix} \hat{a} \cdot \hat{a} & \hat{a} \cdot \hat{b} & \hat{a} \cdot \hat{c} \ \hat{b} \cdot \hat{a} & \hat{b} \cdot \hat{b} & \hat{b} \cdot \hat{c} \ \hat{c} \cdot \hat{a} & \hat{c} \cdot \hat{b} & \hat{c} \cdot \hat{c} \end{vmatrix}
$$

This is a powerful identity. It essentially tells us that the volume is encoded entirely within the dot products of the edges.

### The Calculation: A Dance of Fractions

Since our vectors are unit vectors, the diagonal elements are simply $\hat{a} \cdot \hat{a} = 1, \hat{b} \cdot \hat{b} = 1,$ and $\hat{c} \cdot \hat{c} = 1$. The off-diagonal elements are all $\frac{1}{2}$. Our determinant becomes:

$$
V^2 = \begin{vmatrix} 1 & \frac{1}{2} & \frac{1}{2} \ \frac{1}{2} & 1 & \frac{1}{2} \ \frac{1}{2} & \frac{1}{2} & 1 \end{vmatrix}
$$

Now, let us expand this carefully along the first row:

1. The first term: $1 \cdot (1 \cdot 1 - \frac{1}{2} \cdot \frac{1}{2}) = 1 \cdot (1 - \frac{1}{4}) = \frac{3}{4}$.

2. The second term: $-\frac{1}{2} \cdot (\frac{1}{2} \cdot 1 - \frac{1}{2} \cdot \frac{1}{2}) = -\frac{1}{2} \cdot (\frac{1}{2} - \frac{1}{4}) = -\frac{1}{8}$.

3. The third term: $+\frac{1}{2} \cdot (\frac{1}{2} \cdot \frac{1}{2} - 1 \cdot \frac{1}{2}) = \frac{1}{2} \cdot (\frac{1}{4} - \frac{1}{2}) = -\frac{1}{8}$.

Summing these up, we get:

$$
V^2 = \frac{3}{4} - \frac{1}{8} - \frac{1}{8} = \frac{3}{4} - \frac{2}{8} = \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2}
$$

### The Final Reveal

We have arrived at $V^2 = \frac{1}{2}$. To find the volume $V$, we simply take the square root:

$$
V = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}
$$

Look at how elegant that is! We didn't need to know where the vectors pointed in space; we only needed to know how they interacted with each other. This is the essence of advanced mathematics—finding the underlying structure and letting the algebra flow from there. Keep this Gramian method in your toolkit; it will serve you well in many battles to come.

Select Answer:

Visualized Solution

Geometry of the Parallelepiped

Volume as Scalar Triple Product

The Gramian Determinant

Substituting Dot Products

Determinant Expansion: Row 1

Determinant Expansion: Remaining Terms

Calculating $V^{2}$

Final Volume

The Sigma Insight: Scalar Triple Product

Analyzing the Setup

The Power of the Gramian Determinant

The Calculation

The Final Reveal

Similar Questions

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Match List I with List II:

List-I

List-II

Let $a = \hat{i} - \hat{j}$ , $b = \hat{j} - \hat{k}$ , $c = \hat{k} - \hat{i}$ . If $d$ is a unit vector such that $a \cdot d = 0 = [b c d]$ , then $d$ equals